Fourier transform of $t^n\exp(-\alpha\vert t\vert)$ $(n\geq0)$

Question

I met this Fourier transformation when dealing with bath properties in physics. I would appreciate any clue on this.

Answer 1

Thank @metamorphy for his suggestion. This Fourier transformation can be calculated recursively.

We first split the total integral into two parts $$ \int_{-\infty}^{\infty} dt\ t^n \exp(-\alpha\vert t\vert-i\omega t)= \int_{0}^{\infty} dt\ t^n \exp(-\alpha t-i\omega t) + \int_{-\infty}^{0} dt\ t^n \exp(\alpha t-i\omega t).$$ Denote the first part as $I_n=\int_{0}^{\infty} dt\ t^n \exp(-\alpha t-i\omega t)$ and then take derivative with respect to $\omega$, one can obatin $$I_{n+1}=i\frac{\partial I_n}{\partial \omega}.$$ Starting from $I_0=\frac{1}{\alpha+i\omega}$, the recursive relation yields $$I_n=\frac{n!}{(\alpha+i\omega)^{n+1}}.$$ The same procedure can be done for the second part. The final result reads $$\int_{-\infty}^{\infty} dt\ t^n \exp(-\alpha\vert t\vert-i\omega t)=\frac{n!}{(\alpha+i\omega)^{n+1}}+\frac{(-1)^nn!}{(\alpha-i\omega)^{n+1}}\quad n\in \mathbb{N}$$

Answer 2

Fourier Transform of double decaying exponential with t^n polynomial growth Fourier Transform Pair

  x(t)                         X(j ω)
exp(-α|t|)                    2α/(α^2+ ω^2)
t^n x(t)                      j^n (d^n/dω^n) (X(jw))
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n*************t^n x(t)***********************Xn(j ω)

0 ************* x(t)**********************2α/(ω^2+α^2)

1******* ***** t x(t)*******************-j4ωα/(ω^2+α^2)^2

2********** t^2 x(t)*****************(-16ω^2α)/((ω^2+α^2)^3

n********** t^n x(t)********** (i^n)n!C(ω^n)α /((ω^2+α^2)^(n+1)

All you need for the general solution is to determine constant C.

Hint: look at the pattern generated by this table, can you see a trend is developing?

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